Problem: $\dfrac{ 5e - 3f }{ 6 } = \dfrac{ -3e + 10g }{ -9 }$ Solve for $e$.
Explanation: Multiply both sides by the left denominator. $\dfrac{ 5e - 3f }{ {6} } = \dfrac{ -3e + 10g }{ -9 }$ ${6} \cdot \dfrac{ 5e - 3f }{ {6} } = {6} \cdot \dfrac{ -3e + 10g }{ -9 }$ $5e - 3f = {6} \cdot \dfrac { -3e + 10g }{ -9 }$ Multiply both sides by the right denominator. $5e - 3f = 6 \cdot \dfrac{ -3e + 10g }{ -{9} }$ $-{9} \cdot \left( 5e - 3f \right) = -{9} \cdot 6 \cdot \dfrac{ -3e + 10g }{ -{9} }$ $-{9} \cdot \left( 5e - 3f \right) = 6 \cdot \left( -3e + 10g \right)$ Distribute both sides $-{9} \cdot \left( 5e - 3f \right) = {6} \cdot \left( -3e + 10g \right)$ $-{45}e + {27}f = -{18}e + {60}g$ Combine $e$ terms on the left. $-{45e} + 27f = -{18e} + 60g$ $-{27e} + 27f = 60g$ Move the $f$ term to the right. $-27e + {27f} = 60g$ $-27e = 60g - {27f}$ Isolate $e$ by dividing both sides by its coefficient. $-{27}e = 60g - 27f$ $e = \dfrac{ 60g - 27f }{ -{27} }$ All of these terms are divisible by $3$ Divide by the common factor and swap signs so the denominator isn't negative. $e = \dfrac{ -{20}g + {9}f }{ {9} }$